oldhw7_pdf - mcconnell(kam2342 – oldhomework 07 –...

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mcconnell (kam2342) – oldhomework 07 – Turner – (60230) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Through what potential difference would an electron need to be accelerated for it to achieve a speed of 3 . 2 % of the speed of light (2 . 99792 × 10 8 m / s), starting from rest? Correct answer: 261 . 632 V. Explanation: Let : s = 3 . 2% = 0 . 032 , c = 2 . 99792 × 10 8 m / s , m e = 9 . 10939 × 10 - 31 kg , and q e = 1 . 60218 × 10 - 19 C . The speed of the electron is v = 0 . 032 c = 0 . 032 ( 2 . 99792 × 10 8 m / s ) = 9 . 59336 × 10 6 m / s , By conservation of energy 1 2 m e v 2 = - ( - q e ) Δ V Δ V = m e v 2 2 q e = ( 9 . 10939 × 10 - 31 kg ) × ( 9 . 59336 × 10 6 m / s ) 2 2 (1 . 60218 × 10 - 19 C) = 261 . 632 V . 002 (part 1 of 3) 10.0 points Consider two concentric spherical conducting shells. The smaller shell has radius a = 0 . 2 m and charge 8 μ C on it, while the larger shell has radius b = 3 a and charge 40 μ C on it. Assume that the electric potential V at is zero. The Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . b = 3a a #1 #2 P q 1 q 2 O Determine the electric field E at P, where the distance OP = 2 a . Correct answer: 4 . 49378 × 10 5 N / C. Explanation: Let : a = 0 . 2 m , q 1 = 8 μ C = 8 × 10 - 6 C , q 2 = 40 μ C = 4 × 10 - 5 C , and k = 8 . 98755 × 10 9 N · m 2 / C 2 . Set up a Gaussian surface of radius r = 2 a between the shells. Due to symmetry, the electric field is constant over the surface of the sphere, so the flux is simply Φ = E A . The enclosed charge is q 1 . From Gauss’s Law, Φ = Q encl ǫ 0 E 4 π r 2 = q 1 ǫ 0 , and r = 2 a , so E = q 1 4 π ǫ 0 (2 a ) 2 = k q 1 4 a 2 = ( 8 . 98755 × 10 9 N · m 2 / C 2 ) 8 × 10 - 6 C 4 (0 . 2 m) 2 = 4 . 49378 × 10 5 N / C . 003 (part 2 of 3) 10.0 points Find the electric potential V at point P . 1. 1 4 π ǫ 0 parenleftBig q 1 3 a + q 2 2 a parenrightBig 2. 1 4 π ǫ 0 parenleftbigg q 1 + q 2 2 a parenrightbigg 3. 1 4 π ǫ 0 parenleftBig q 1 2 a + q 2 3 a parenrightBig correct
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mcconnell (kam2342) – oldhomework 07 – Turner – (60230) 2 4. 1 4 π ǫ 0 parenleftBig q 1 2 a + q 2 a parenrightBig 5. 1 4 π ǫ 0 parenleftbigg q 1 + q 2 3 a parenrightbigg 6. 1 4 π ǫ 0 parenleftBig q 1 a + q 2 3 a parenrightBig 7. 1 4 π ǫ 0 parenleftBig q 1 a + q 2 2 a parenrightBig 8. 9. 0 10. 1 4 π ǫ 0 parenleftbigg q 1 + q 2 a parenrightbigg Explanation: The potential due to a spherical charge dis- tribution is the same as the potential due to a point charge at its center, if V at is zero (otherwise we would add a constant). Due to q 1 , V 1 = k q 1 r = 1 4 π ǫ 0 q 1 2 a The outer sphere (#2) requires a little more thought. Just outside the surface, the poten-
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